Chapt12

Physical Properties of SolutionsChapter 12

Dr. Ali Bumajdad

Chapter 12 Topics

• Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization•Solution Process•Concentrations (%mass, Mole fraction, molality, molarity, ppm)•Factors affecting Solubility (T and P)•Colligative properties of nonelectrolyte solution: 1)Vapour pressure lowering 2)B.P. Elevation 3) F.P. Depression 4) Osmotic Pressure ()•Using Colligative properties to determine molar mass•Colligative properties of electrolyte solution

Physical Properties of Solutions

Dr. Ali Bumajdad

•Solution: homogenous mixture of 2 or more substances

•Solute: is(are) the substance(s) present in the smaller amount(s)

•Solvent: is the substance present in the larger amount

Solution not necessarily liquid, it could be gas or solid

Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization

• Saturated solution: contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

• Unsaturated solution: contains less solute than the solvent has the capacity to dissolve at a specific temperature.

• Supersaturated solution: contains more solute than is present in a saturated solution at a specific temperature (not stable).

Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate

• Crystallization: dissolved solute comes out of solution and forms crystals.

Solute + solvent Solutionsolvation

crystallization

Three types of interactions in the solution process:• solvent-solvent interaction (endothermic)• solute-solute interaction (endothermic)• solvent-solute interaction (endothermic or exothermic)

Hsoln = H1 + H2 + H3

Solution Process

Hsoln>0, solution may form and

maynotHsoln<0, solution

favourable

“like dissolves like”

Two substances with similar intermolecular forces are likely to be soluble in each other.• non-polar molecules are soluble in non-polar solvents

CCl4 in C6H6

• polar molecules are soluble in polar solvents

C2H5OH in H2O

• ionic compounds are more soluble in polar solvents

NaCl in H2O or NH3 (l)

When solute-solute or solvent-solvent interaction are much more than solute-solvent interaction a solution will not form

•Miscible: two liquids completely soluble in each other in all proportions

•Solvation: ion or molecules surrounded by solvent molecules

Q) Can we form a solution of octane (C8H18) in water?

No because: - octane not polar while water is polar-In this case H3 is small while H1 and H2 large +ve

hence, Hsoln is large positive

Sa Ex. :Predict whether each of the following substances is more likely to dissolve in CCl4 or in Water: C7H16, Na2SO4, HCl, I2.

(a) Br2 in C6H6

(b) KCl in NH3

(c) CH2O in H2O

Concentration Units•Concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

1) Percent by Mass

= x 100%mass of solutemass of solution

2) Mole Fraction (X)

% by mass = x 100%mass of solutemass of solute + mass of solvent

(1)

XA = moles of A

sum of moles of all components(2)

Concentration Units Continued

3) Molarity (M)

4) Molality (m)

M =moles of solute

liters of solution

(3)

m =moles of solute

mass of solvent (kg)

(4)

Concentration Type Equation Equation Unit Mass Percentage 100

solutionofmasstotal

solutionincomponentofmasscomponentof%Mass 100

.......mm

m%m

ba

a No unit (%)

Mole Fraction molesof.nototal

componentofmolesof.nocomponentoffractionMole

.......nn

nX

ba

aa No unit

Molarity literinsolutionofvolumetotal

soluteofmolesof.noMolarity tot

LV

nM

mol/l or M

Molality Kginsolventofmass

soluteofmolesof.noMolality solvent

kgm

nm

mol/kg or m

1) n= massM.m

2) D= massVol

Sa. Ex.:A solution is made by dissolving 13.5 g glucose (C6H12O6) in

0.100 kg of water. What is the mass percentage of solute in this solution?

Sa. Ex. (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50. g of water.(b) A commercial bleaching solution contains 3.62 mass% sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

11.9%

2.91%

90.5g

Sa. Ex.a) A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 ml of water. Calculate the molality of glucose in the solution.

b) What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)

c) A solution of hydrochloric acid contains 36% HCl by mass (i) Calculate the mole fraction of HCl in the solution. (ii) Calculate the molality of HCl in the solution

0.966 m

0.670 m

XHCl= 0.22

15 m

Q) What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solutionAssume 1 L of solution:

(1) mass of solute: (nMm=mass) 5.86 moles ethanol contains 270 g ethanol (2) mass of solution: solution density solution volume927 g of solution

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

Sa. Ex. A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has a density of 0.876 g/ml. Calculate the molarity of the solution.

Sa. Ex. A solution containing equal masses of glycerol (C3H8O3, Mm=92gmol-1) and water has a density of 1.10 g/ml. Calculate:(a) The molality of glycerol(b) The mole fraction of glycerol(c) The molarity of glycerol in the solution.

0.21 M

10.9 m0.1635.97 M

(1) Temperature

a) Solid solubility and temperature

Solubility increases with increasing temperature

Solubility decreases with increasing temperature

•Factors affecting Solubility (T and P, and nature of solute and solvent)

Like dissolve like

Usually s T For solid

•Fractional crystallization: is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL of water at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

(b) Gas solubility and temperature

Solubility usually decreases with

increasing temperature

Usually s 1 For gas

T

(1) Temperature

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

c is the concentration (M) of the dissolved gas

P is the pressure of the gas over the solution (vapor or partial pressure)

k is a constant for each gas (mol/L•atm) that depends only on temperature

low P

low c

high P

high c

(2) Pressure

a) Gas solubility and Pressure

Vapour Pressure: pressure of gas above the

liquid

(there are some exceptionswhen the gas react with water) c = kP

(5)Henry’s law

Sample Ex. :Calculate the concentration of CO2 in soft drink (partial pressure of CO2 = 4.0 atm, and the Henery's law constant for CO2 in water is 3.1 10-2 mol/L atm.

c = kP(5)

=0.12 mol/l

Colligative Properties of Nonelectrolyte Solutions

•Colligative properties: properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

1) Vapor-Pressure Lowering

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

(e.g. adding salt and ethylene glycol reduce F.P.)

P1 = X1 P 10 (6)

Raoult’s law

For nonvolatile solute (no measurablevapour pressure)

Where:

Adding nonvolatile solute to a solvent always lower the vapour pressure

P 1 = vapor pressure of solution after adding solute

If the solution contains only one solute:X1 = 1 – X2

X2 = mole fraction of the solute

P 1 = P 10(1 - X2)

The decrease in vapour pressure (P) directly proportion to solute concentration (not to the nature of solute)

P 10 - P1 = P = X2 P 1

0 (7)

PA = XA P A0

PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

Ideal Solution

When both components are volatile:1) Ideal behaviour

PT is greater thanpredicted by Raoults’s law

PT is less thanpredicted by Raoults’s law

ForceA-B

ForceA-A

ForceB-B< &

ForceA-B

ForceA-A

ForceB-B> &

When both components are volatile: 2) Nonideal behaviour

Fractional Distillation ApparatusUsed to separate two liquids from each other based on their boiling point

P 10 - P1 = P = X2 P 1

0 (7)

P1 = X1 P 10 (6) Raoult’s Law

X1 = n1

n1+n2

(2)

X2=1-X1

Vapour pressure of water at 30 °C = 31.82mmHg

Sa. Ex. Glycerin (C3H8O3) is a nonvolatile nanelectrolyte with a density of 1.26 g/ml at 25 °C. Calculate the vapor pressure at 25 °C of a solution madeby adding 50.0 ml of glycerin to 500.0 ml of water. The vapor pressure of purewater at 25 °C is 23.8 torr.

Sa. Ex. The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?

23.2 torr

0.290

2) Boiling-Point Elevation

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m) for a given solvent (depend only on solvent)

Tb0 Tb

Tb = Kb m(8)

Tb = Tb – T b0By the addition of nanvolatile solute

3) Freezing-Point Depression

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m) for a given solvent (depend only on solvent)

Tf = T f – Tf0

Tf = Kf m(9)

Tf Tf0

By the addition solute

The decrease in Tf upon addition of solute is due to the more energy needs to be removed to form order solution than to formorder solvent (solution is more disorder)

The solid that separate when a solution freeze is the pure solvent onlythat why the V.P. of solid is the same as that for pure liquid (see the phase diagram)

De-icing of airplanes

Q) What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g/mol.

m =moles of solute

mass of solvent (kg)= = 2.41 m

3.202 kg solvent

478 g

62.01 g

(Kf water = 1.86 0C/m)

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

Tf = Kf m(9)

Sa. Ex. Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing pointof a 25.0 mass% solution of ethylene glycol in water.

Sa. Ex. Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance foundin the leaves of eucalyptus trees.

(Kb water=0.52 and Kf water = 1.86 °C/m)

Tb = Kb m(8)

Tf = Kf m(9)

Tb = Tb0 +Tb Tf = Tf

0 -Tb

102.8 °C -10.0 °C

4) Osmotic Pressure ()

•Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

•Semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

• Osmotic pressure () is the pressure required to stop osmosis.

dilutemore

concentrated

Cucumber in concentrated brine lose water Pickle

HighP

LowP

M is the molarity of the solution

R is the gas constant (0.0821 L atm/mol K)

T is the temperature (in K)

= MRT(10)Osmotic pressure

in unit atm

Colligative property = depend on concentration only

•Isotonic: the two solutions are of equal concentration and, hence, they have the same osmotic pressure

•Hypotonic: the solution of lower concentration

• When the concentrations of the two solution are equal and, hence, have the

same osmotic pressure, the solutions called isotonic

•When concentrations of the two solution are different, the lower concentration

called hypotonic and the higher concentration called hypertonic

•Hypertonic: the solution of higher concentration

A cell in an:

A cell inisotonicsolution

A cell inhypotonic

solution

A cell inhypertonic

solution

the two solution are equal in concentration

the outer solution is more concentrated than the

solution in the cell

the outer solution is lessconcentrated than the

solution in the cell

= MRT(10)

1.23 M

Sa. Ex.:The average osmotic pressure of blood is 7.7 atm at 25 °C.what concentration of glucose (C6H12O6) will be isotonic with blood?

Sa. Ex.:What is the osmotic pressure at 20 °C of a 0.0020 M sucrose (C12H22O11) solution?

0.31 M

0.048 atm

•Using Colligative Properties to determine molar mass

Tb = Kb m(8)

m= Tb

Kb

m = nmass of solvent in Kg

m = mass in g

M.m. (mass of solvent in Kg)

Tf = Kf m(9)

m= Tf

Kf

m = nmass of solvent in Kg

m = mass in g

M.m. (mass of solvent in Kg)

M.m = mass in g

m (mass of solvent in Kg)

(11)M.m = mass in g

m (mass of solvent in Kg)

(11)

B.P. Elevation F.P. Depression

M = nVL

M = mass in gM.m. (VL)

M.m = mass in gM (VL)

(12)

= MRT(10)

M = RT

Osmotic Pressure

M.m = (mass in g) R T (VL)

(13)

Kf of benzene = 5.12 °C/m

m= Tf

Kf

M.m = mass in g

m (mass of solvent in Kg)

(11)

127 g/mol

C10H8

1 atm = 760 mmHg

6.51 104 g/mol

Sa. Ex.: Coamphor (C10H16O) melts 1t 179.8 °C, and it has a particularly large freezing-point-depression constant Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. what is the molar mass of the solute?

Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

(8)

(9)

(10)

(7)

Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions

•Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

0.1 m NaCl solution 0.2 m ions in solution

nonelectrolytesNaCl

CaCl2

i should be

12

3

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

Colligative Properties of Electrolyte Solutions

Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

(8b)

(9b)

(10b)

= i MRT

Sa. Ex.: List the following solutions in the order of their expected freezing points: 0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22O11.

Sa. Ex: Which of the following solutes will produce the largest increase in boiling point upon addition to 1 Kg of water: 1 mol Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)?